Find :


Find $\frac{d y}{d x}$ :

$2 x+3 y=\sin x$


The given relationship is

Differentiating this relationship with respect to x, we obtain

$\frac{d}{d x}(2 x+3 y)=\frac{d}{d x}(\sin x)$

$\Rightarrow \frac{d}{d x}(2 x)+\frac{d}{d x}(3 y)=\cos x$

$\Rightarrow 2+3 \frac{d y}{d x}=\cos x$

$\Rightarrow 3 \frac{d y}{d x}=\cos x-2$

$\therefore \frac{d y}{d x}=\frac{\cos x-2}{3}$

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